Problem 70 - A Graphical Approach to College Algebra

Question

Sketch a graph of a quadratic function that satisfies each set of given conditions. Use symmetry to label another point on your graph. Vertex \((5,6) ;\) through \((1,-6)\)

Step-by-Step Solution

Answer

The quadratic function is \( f(x) = -\frac{3}{4}(x-5)^2 + 6 \). Symmetric point: \((9, -6)\).

Step 1: Identify the Vertex Form

A quadratic function in vertex form is expressed as: \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. For this problem, the vertex given is \((5, 6)\). Thus, the equation becomes \( f(x) = a(x-5)^2 + 6 \).

Step 2: Substitute the Given Point

Use the point \((1, -6)\) that the parabola passes through to find \(a\). Substitute \( x = 1 \) and \( f(x) = -6 \) into the vertex form: \( -6 = a(1-5)^2 + 6 \).

Step 3: Solve for 'a'

Simplify the equation from Step 2: \( -6 = a(16) + 6 \). Isolate \(a\): \( -12 = 16a \Rightarrow a = -\frac{12}{16} \Rightarrow a = -\frac{3}{4} \).

Step 4: Write the Equation of the Quadratic Function

Substitute \(a\) back into the vertex form: \( f(x) = -\frac{3}{4}(x-5)^2 + 6 \). This is the equation of the parabola.

Step 5: Use Symmetry to Find Another Point

Since the vertex is \((5, 6)\), the axis of symmetry is \(x = 5\). Since the point \((1, -6)\) is 4 units to the left of the vertex, find a symmetric point that is 4 units to the right: \((9, -6)\).

Key Concepts

Parabola Vertex Form

The vertex form of a quadratic function is incredibly useful when graphing parabolas. This form is written as \( f(x) = a(x - h)^2 + k \), where \((h, k)\) represents the vertex of the parabola. The vertex is the highest or lowest point on the parabola, depending on whether it opens upwards or downwards. Knowing the vertex form helps create the basic shape of the parabola quickly and identifies the vertex immediately.
In the given exercise, the vertex is \((5, 6)\), so we start by plugging these values into the vertex form, forming \( f(x) = a(x - 5)^2 + 6 \). At this stage, \(a\) is unknown, but the equation provides the skeleton of our parabola structure.

Axis of Symmetry

The axis of symmetry is a vertical line that runs through the vertex of the parabola. This line helps divide the parabola into two equal mirror halves. Knowing this line is crucial as it allows us to assess the symmetry and find other symmetric points on the graph.
In this problem, the axis of symmetry is \(x = 5\) because the given vertex is \((5, 6)\). This means any point along the parabola has a mirrored counterpart that is equidistant from this line of symmetry. Understanding this symmetry helps when plotting additional points, ensuring that your graph of the parabola is accurate and balanced.

Finding Coefficients

Finding the coefficient \(a\) in the vertex form helps us determine the vertical stretch or compression of the parabola, as well as the direction it opens—upward or downward. To find \(a\), we need another point through which the parabola passes. Here, the point given is \((1, -6)\).
We substitute this point into the vertex form equation: \(-6 = a(1-5)^2 + 6\) and solve for \(a\). Simplifying gives \(-6 = 16a + 6\), or \(-12 = 16a\). Solving gives \(a = -\frac{3}{4}\). This value indicates that the parabola opens downward and is vertically compressed.

Graphing Parabolas

Graphing a parabola begins with plotting its vertex on the coordinate plane. Next, using the axis of symmetry, you can plot additional symmetric points. Start by plotting both the vertex \((5, 6)\) and the known point \((1, -6)\).
To find another symmetric point, measure the distance from the vertex to the known point along the x-axis. Since \((1, -6)\) is 4 units to the left of the axis of symmetry \(x=5\), its symmetric counterpart will also be 4 units to the right, at \((9, -6)\).

• Plot the vertex and points \((1, -6)\) and \((9, -6)\).
• Draw the parabola opening downwards, passing through these points.

This ensures the graph reflects a properly symmetric parabola.